Electric Field from Point Charges

In summary: However, if you can't explain what you did wrong or how it could impact the result, then you should not include it in your answer.
  • #1
tristanm
7
0

Homework Statement


Two point charges (q1 = -2.6uC and q2 = 8.6uC) are fixed along the x-axis, separated by a distance d = 9.7cm. Point P is located at (x,y) = (d,d)

1. What is Ex(P), the x-component of the electric field produced by q1 and q2 at point P?
2. What is Ey(P), the value of the y-component of the electric field produced by q1 and q2 at point P?
3.A third charge is now positioned along the y-axis at distant d from and above q1 on (x,y) = (0,d). What is the Ex(P) of the field produced by all 3 charges at point P?
4. Suppose all are doubled, how will the electric field change?
5. How would you change Q1 in order to make the eletric field at point P equal to zero?


Homework Equations


E = (kq)/r^2
Exq1 = ((kq)/r^2)cos(45)
Exq2 = 0
Eyq1 = ((kq/r^2)sin(45)
Eyq2 = (kq)/r^2
C^2 = A^2 + B^2

Image for 1 and 2
h2_pointA.png


Image for 3, 4, and 5
h2_pointC.png


The Attempt at a Solution


1. First off, I drew the triangle of q1-q2-P and labelled all d values as 0.097m, used E=(kq)/r^2 to find the magnitude of E at point P a distance r = sqrt(2d^2) and then multiply that value by cos(45) to find that Exq1 is equal to -120608.8429 N/C by having q = 2.6e-6 C, k = 8.987551e9 and r = (which apparently is wrong)

2. Ey = Eyq1 + Eyq2, where Eyq1 = ((kq1)/r^2) * sin45 and Eyq2 = (kq2)/d^2. For Eyq1 I calculated -120451.7144 N/C and for Eyq2 I calculated 8214787.82 N/C. The sum of the two is 8094336.106 N/C which again, is apparently wrong.

I have not yet attempted 3,4 or 5 because I recognize after the first two I'm clearly doing something wrong in my steps.
 
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  • #2
First off, you are using way too many significant figures. The data you are given only have 2 significant figures, so your answers should have no more than that.

Now, looking at part 1, your procedure seems correct, so you should double check your arithmetic. Make sure whatever you are using to calculate the cosine is expecting degrees and not radians.
 
  • #3
I realize that the sig figs aren't accurate. I use smartphysics for my homework assignments and it is notorious for not accepting correct sig figs. My problem was that I did not actually do the arithmetic correctly.

Thank you
 
  • #4
Even if your program or calculator gives you many digits, you should only report those that are significant. As for arithmetic mistakes, we all make them.
 
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  • #5



Hello, thank you for your question. I would like to address your concerns and provide a response to your content.

1. To find the x-component of the electric field at point P, we can use the equation E = (kq)/r^2, where k is the Coulomb's constant, q is the charge, and r is the distance between the point charge and point P. In this case, we have two point charges, q1 and q2, separated by a distance d. Therefore, the magnitude of the electric field at point P due to q1 is given by E1 = (kq1)/r1^2, where r1 = d/sqrt(2) is the distance between q1 and P. Similarly, the magnitude of the electric field at point P due to q2 is given by E2 = (kq2)/r2^2, where r2 = d/sqrt(2) is the distance between q2 and P. The x-component of the electric field at point P is then given by the vector sum of E1 and E2, which can be calculated using the Pythagorean theorem as Ex(P) = sqrt(E1^2 + E2^2) * cos(45). Substituting the values of q1, q2, and d, we get Ex(P) = -1.206 * 10^5 N/C.

2. Similarly, to find the y-component of the electric field at point P, we can use the same approach as in part 1, but this time using the sine function instead of cosine. The y-component of the electric field at point P due to q1 is given by Ey1 = (kq1)/r1^2 * sin(45), and the y-component of the electric field at point P due to q2 is given by Ey2 = (kq2)/r2^2. The y-component of the electric field at point P is then given by the vector sum of Ey1 and Ey2, which can be calculated using the Pythagorean theorem as Ey(P) = sqrt(Ey1^2 + Ey2^2). Substituting the values of q1, q2, and d, we get Ey(P) = 8.094 * 10^6 N/C.

3. To find the x-component of the electric field at point P
 

What is an electric field?

An electric field is a physical quantity that describes the influence a charged particle has on other charged particles in its vicinity. It is a vector quantity, meaning it has both magnitude and direction.

What is a point charge?

A point charge is a hypothetical object that has a single positive or negative charge concentrated at a single point in space. It is often used in physics equations to simplify calculations.

How is the electric field from a point charge calculated?

The electric field from a point charge is calculated using Coulomb's law, which states that the electric field strength is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charge and the point at which the field is being measured.

What is the direction of the electric field from a positive point charge?

The direction of the electric field from a positive point charge is radially outward, meaning it points away from the charge in all directions. This is because positively charged particles repel each other.

How does the electric field from multiple point charges add up?

The electric field from multiple point charges can be calculated by adding the individual electric fields from each charge vectorially. This means that the direction and magnitude of the electric field at a specific point is determined by the vector sum of all the individual electric fields at that point.

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